Question: $\overline{AB}$ = $4\sqrt{5}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $4\sqrt{5}$ $?$ $ \sin( \angle ABC ) = \frac{ \sqrt{5}}{5}, \cos( \angle ABC ) = \frac{2\sqrt{5} }{5}, \tan( \angle ABC ) = \dfrac{1}{2}$
$\overline{AB}$ is the hypotenuse $\overline{AC}$ is opposite to $\angle ABC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle ABC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{4\sqrt{5}} $ $ \overline{AC}=4\sqrt{5} \cdot \sin( \angle ABC ) = 4\sqrt{5} \cdot \frac{ \sqrt{5}}{5} = 4$